Question

In the adjacent figure, $\angle B< \angle A$ and $\angle C< \angle D$ . Show that $AD< BC$.

Answer 1

Hint: To prove $AD< BC$, we are going to use the property which says that the side opposite to the larger angle is longer and we will apply this property on to the angles A and angle B and then sides opposite to those angles. Also, we are going to apply this property onto the angles C and angle D and sides opposite to those angles.

Complete step by step solution:
The figure given in the above problem is as follows:

It is given that $\angle B<\angle A$ so side opposite to angle B is AO and side opposite to angle A is BO then we are going to apply the property that side opposite to larger angle is longer side so angle A is the larger angle as compared to angle B so side opposite to angle A (i.e. BO) is longer than side AO which is opposite to angle B.
$AO< BO$ ………… (1)
Similarly, we are going to apply the same property on angles C and D. It is given that $\angle C<\angle D$ so the side opposite to angle D (i.e. OC) is longer than side opposite to angle C (i.e. OD).
$OD< OC$ ……… (2)
Adding (1) and (2) we get,
$AO+OD< BO+OC$
 From the above figure, we can see that:
$\begin{align}
  & AO+OD=AD; \\
 & BO+OC=BC \\
\end{align}$
So, using the above relations in the above inequality we get,
$AD Hence, we have shown that $AD< BC$.

Note: Just like in the above problem, we have shown that the side opposite to the larger angle is longer. Similarly, sides opposite to equal angles are equal. Also, the side opposite to the shorter angle is small in size as compared to the side opposite to the larger angle.